At a party everyone shakes hands, 66 hand shakes occur, how many people are at the party?

12

You can suppose there are n people in the room and think of them in a row. The first one has to shake hands with (n-1) people (because he doesn't have to shake hands with himself). The second one has already shaken hands with the first one, so he has (n-2) shakes remaining... and so on. So you have to sum: (n-1)+(n-2)+(n-3)+...+1= (n/2)*(n-1) Then you have to solve (n/2)*(n-1)=66 and you get n=12.

n(n-1)/2=66 so n=12

You can think of this problem as a combination problem: (N choose 2) = 66, and then solve for N. That is, N! / (2! (N-2)! ) = 66. Simplifying this equation leads to N(N-1)/2 = 66, and the integer solution to this problem is N = 12.

66=x!/((x-2)!*2) which gives 12

requires explanation

12

sum of the series 66= n/2( 2*1 + (n-1)*1) n=12